thermodynamics (2013 new edition) copy
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THERMODYNAMICS Part 1By. Engr. Yuri G. Melliza
Terms & Definition
Properties of Fluids
Forms of Energy
Law of Conservation of Mass
Law of Conservation of Energy
(First Law of Thermodynamics)
Ideal Gas
Pure Substance
Processes of Fluids
Zeroth Law of Thermodynamics
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Thermodynamics is a science that deals with energy transformation or conversion of one form of energy to another form
Therme – “Heat”
Dynamis – “Strength”
System: A portion in the universe, an Atom, a Galaxy, a certain quantity of matter, or a certain volume in space in which one wishes to study. It is a region enclosed by a specified boundary
that may Imaginary, Fixed or Moving.
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System
Surrounding or Environment
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Open System: A system open to matter flow.
Example: Internal Combustion Engine (ICE)
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Closed System: A system close to matter flow.
Example: Piston - in - cylinder
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Working Substance (Working Fluid): A fluid (Liquid or Gas) responsible for the transformation of energy.
Example: air in an air compressor Air and fuel mixture in an internal combustion
engine
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Pure Substance: A substance that is homogeneous in nature and is homogeneous.
Example : Water
Phases of a Substance A phase refers to a quantity of matter that is homogeneous throughout in both chemical composition and physical structure. Solid Liquid Gas or Vapor
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Specific Terms To Characterized Phase Transition
SOLIDIFYING OR FREEZING - Liquid to Solid
MELTING - Solid to Liquid VAPORIZATION - Liquid to Vapor CONDENSATION - Vapor to Liquid SUBLIMATION - a change from solid
directly to vapor phase without passing the liquid phase.
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Mass : It is the absolute quantity of matter in it.
m - mass in kg
Acceleration : it is the rate of change of velocity with respect to time t.
a = dv/dt m/sec2
Velocity: It is the distance per unit time.
v = d/t m/sec
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Force - it is the mass multiplied by the acceleration.F = ma/1000 KN
1 kg-m/sec2 = Newton (N) 1000 N = 1 Kilo Newton (KN)
Newton - is the force required to accelerate 1 kg mass at the rate of 1 m/sec per second.
1 N = 1 kg-m/sec2
From Newton`s Law Of Gravitation: The force of attraction between two masses m1 and m2 is given by the equation:
Fg = Gm1m2/r2 NewtonWhere: m1 and m2 - masses in kg
r - distance apart in meters G - Gravitational constant in N-m2/kg2
G = 6.670 x 10 -11 N-m2/kg2
WEIGHT - is the force due to gravity.W = mg/1000 KN
Where: g - gravitational acceleration at sea level, m/sec2 g = 9.81 m/sec2
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3mkg
Vm ρ
PROPERTIES OF FLUIDS
Where: - density in kg/m3
m - mass in kg V – volume in m3
Specific Volume () - it is the volume per unit mass or the reciprocal of its density.
kgm
mV
3
υ
kgm
3
ρ1 υ
Density () - it is the mass per unit volume.
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Specific Weight () - it is the weight per unit volume.
3
3
mKN
mKN
1000V
mgγ
V
Wγ
Where: - specific weight in KN/m3
m – mass in kg V – volume in m3
g – gravitational
At standard condition:g = 9.81 m/sec2
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Specific Gravity Or Relative Density (S):FOR LIQUIDS: Its specific gravity or relative density is equal tothe ratio of its density to that of water at standard temperature and pressure.
w
L
w
L
L γ
γ
ρ
ρS
FOR GASES: Its specific gravity or relative density is equal to theratio of its density to that of either air or hydrogen at some specified temperature and pressure
ah
GG ρ
ρS
Where at standard condition: w = 1000 kg/m3
w = 9.81 KN/m3
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Temperature: It is the measure of the intensity of heat in a body.Fahrenheit Scale:
Boiling Point = 212 FFreezing Point = 32 F
Centigrade or Celsius Scale:Boiling Point = 100 CFreezing Point = 0 C
Absolute Scale: R = F + 460 (Rankine) K = C + 273 (Kelvin)
32F8.1F8.132F
C
Conversion
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Pressure: It is the normal component of a force per unit area.
KPa or 2m
KN
AF
P
Where: P – pressure in KN/m2 or KPa F – normal force in KN A – area in m2
1 KN/m2 = 1 KPa (KiloPascal) 1000 N = 1 KN
If a force dF acts on an infinitesimal area dA, the intensity of Pressure is;
KPa or 2m
KN
dAdF
P
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Pascal’s Law: At any point in a homogeneous fluid at rest the pressures are the same in all directions:
y
x
z
A
BC
P1A1
P2A2
P3A3
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Fx = 0 From Figure:P1A1 - P3 A3sin = 0
P1A1 = P3A3sin Eq.1 P2A2 - P3A3cos = 0
P2A2 = P3A3 cos Eq.2 sin = A1/A3
A1 = A3sin Eq.3cos = A2/A3
A2 = A3cos Eq.4 substituting eq. 3 to eq. 1 and eq.4 to eq.2
P1 = P2 = P3
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Atmospheric Pressure (Pa):It is the average pressure exerted by the atmosphere. At sea level
Pa = 101.325 KPa = 0.101325 MPa= 1.01325 Bar = 760 mm Hg = 10.33 m of water= 1.033 kg/cm2
= 14.7 lb/in2
Pa = 29.921 in Hg = 33.88 ft. of water
100 KPa = 1 Bar 1000 KPa = 1 MPa
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Absolute and Gauge PressureAbsolute Pressure: It is the pressure measured referred to absolute zero using absolute zero as the base.Gauge Pressure: it is the pressure measured referred to the existing atmospheric pressure and using atmospheric pressure as the base.
Pgauge – if it is above atmosphericPvacuum – negative gauge or vacuum if it is below
atmospheric
Barometer: An instrument used to determine the absolute pressure exerted by the atmosphere
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Atmospheic pressure (Pa)
Absolute Zero
Pvacuum
Pgauge
Pabsolute
Pabsolute
Pabs = Pgauge + PaPabs = Pvacuum - Pa
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VARIATION OF PRESSURE
PA
(P + dP)AW
dh
F = 0(P + dP)A - PA - W = 0 PA + dPA - PA - W = 0dPA - W = 0 or dPA = W Eq. 1but : W = dV dPA = - dV
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where negative sign is used because distance h is measured upward and W is acting downward.
dV = Adh then dPA = -Adh, therefore
dP = - dh (Note: h is positive when measured upward and negative if measured downward)
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MANOMETERSManometer is an instrument used in measuring gage pressure in length of some liquid column.1. Open Type Manometer : It has an atmospheric surface and is capable in measuring gage pressure.2. Differential Type Manometer : It has no atmospheric surface and is capable in measuring differences of pressure.
Open TypeOpen end
Manometer Fluid
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Differential Type
Fluid A
Fluid B
Fluid C
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ENERGY FORMSWork: It is the force multiplied by the displacement in the direction of the force.
W =∫Fdx KJ-W - indicates that work is done on the system+W - indicates that work is done by the system.
Heat: It is a form of energy that crosses a system's boundary, because of a temperature difference between the system and the surrounding.
Q - Heat KJ +Q - indicates that heat is added to the system-Q - indicates that heat is rejected from the system.
Internal Energy: It is the energy acquired due to the overall molecular interaction, or the total energy that a molecule has.
U = mu KJU - total internal energy KJu - specific internal energy KJ/kgU- change of internal energy
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Flow Energy Or Flow Work: It is the energy required in pushing a fluid usually into the system or out from the system.
System orControl Volume
P1
P2
A1
A2
L1
L2
Ef1 = F1L1
F1 = P1A1
Ef1 = P1A1L1
A1L1 = V1
Ef1 = P1V1
Ef2 = F2L2
F2 = P2A2
Ef2 = P2A2L2
A2L2 = V2
Ef2 = P2V2
Ef = Ef2 – Ef1
Ef = P2V2 – P1V1
Ef = PV
PV = P2V2 - P1V1 KJ
P = P22 - P11 m3/kg
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Where: P – pressure in KPaV – volume in m3
- specific volume in m3/kgEf = PV – Flow energy or flow work
Kinetic Energy: It is the energy acquired due to the motion of a body or a system.
1 2
m mF
d x
dxFdKE
dxFKE
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KJ/kg
2(1000)ΔKE
KJ 2(1000)
mΔKE
dtdv
dx1000
mKE
dtdv
1000m
1000ma
F
FdxKE
vv
vv2
1
2
2
2
1
2
2
2
1
21000m
ΔKE
dvv1000m
ΔKE
dtdx
dv1000m
ΔKE
vv2
1
2
2
2
1
2
1
Where: m – mass , kg v – velocity , m/sec
kg
KJ
10002vv
KE
KJ 10002
vvmKE
21
22
21
22
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Potential Energy: It is the energy required by virtue of its configuration or elevation.
m
m
dZ
Reference Datum
kgKJ
1000
ZZgPE
KJ 1000
ZZmgPE
dZ1000mg
PE
dZWPE
12
12
Where:W – WorkQ – HeatU – Internal EnergyPV – Flow Energy or flow workKE – Kinetic EnergyPE – Potential Energy
Note: +Z – if measured upward - Z –if measured downward
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Law of Conservation of Mass
Mass is indestructible: In applying this law we must except nuclear processes during which mass is converted into energy.
The verbal form of the law is:Mass Entering - Mass Leaving = Change of Mass stored in the systemIn equation Form:
m1 - m2 = m
1 2
m1 m2m = 0
ab c
d
For a steady-state, steady-flow system m = 0, thereforem1 - m2 = 0 or m1 = m2
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For one dimensional flow, where1 = 2 = Let m1 = m2 = m
Continuity Equation:
υ
AvAvρm
Where:m - mass flw rate in kg/sec - density in kg/m3
- specific volume inm3/kgA - cross sectional area in m2
v - velocity in m/sec
υυυ
ρ
ΑvvΑvΑΑvvΑρvΑρ
mmm
2
22
1
11
222111
21
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Zeroth Law of ThermodynamicsIf two bodies are in thermal equilibrium with a third body, they are in thermal equilibrium with each other, and hence their temperatures are equal.
Specific Heat or Heat Capacity: It the amount of heat required to raise the temperature of a 1 kg mass of a substance 1C or 1K.
tmCTmCQ
m; gConsiderin
Cdt CdT dQ
C;constant ForK-kg
KJ or
C-kgKJ
dtdQ
dTdQ
C
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SENSIBLE HEAT: The amount of heat per unit mass that must be transferred (added or remove) when a substance undergoes a change in temperature without a change in phase.
Q = mC(t) = mC(T)where: m - mass , kgC - heat capacity or specific heat, KJ/kg-C or KJ/kg- Kt - temperature in CT - temperature in K
HEAT OF TRANSFORMATION: The amount of heat per unit mass that must be transferred when a substance completely undergoes a phase change without a change in temperature.
Q = mL
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A. Heat of Vaporization: Amount of heat that must be added to vaporize a liquid or that
must be removed to condense a gas.Q = mL
where L - latent heat of vaporization, KJ/kgB. Heat of Fusion : Amount of heat that must be added to melt a solid or that must be removed to freeze a liquid.
Q = mLwhere L - latent heat of fusion, KJ/kg
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THE FIRST LAW OF THERMODYNAMICS (The Law of Conservation of (Energy)
“Energy can neither be created nor destroyed but can only be converted from one form to another.”Verbal Form:
Energy Entering – Energy Leaving = Change of Energy stored in the systemEquation Form:
E1 – E2 = Es1. First Corollary of the First Law: Application of first Law to a Closed System
U
Q
WFor a Closed System (Non FlowSystem), PV, KE and PE are negligible, therefore the changeof stored energy Es = U
Q – W = U 1Q = U + W 2
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By differentiation:dQ = dU + dW 3
where:
dQ Q2 – Q1
dW W2 – W1
Work of a Closed System (NonFlow)
P
V
W = PdV
P
dV
5 Eq. dVPUQ
4 Eq. dVPdUdQ
3 Eq. From
dVPdW
dVPW
dVAdx dxPAW
PAF dxFW
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2. Second Corollary of the First Law: Application of First Law to an Open System
System orControl volume
Datum Line
Q
W
1
2
U1 + P1V1 + KE1 + PE1
U2 + P2V2 + KE2 + PE2
For an Open system (Steady state, Steady Flow system) Es = 0, thereforeE1 – E2 = 0 or E1 = E2 orEnergy Entering = Energy Leaving
Z1
Z2
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U1 + P1V1 + KE1 + PE1 + Q = U2 + P2V2 + KE2 + PE2 + W 1Q = (U2 – U1) + (P2V2 – P1V1) + (KE2 – KE1) + (PE2 – PE1) + W 2Q = U + (PV) + KE + PE + W 3By differentiationdQ = dU + d(PV) + dKE + dPE + dW 4But dQ Q2 – Q1 and dW W2 – W1
Enthalpy (h)h = U + PVdh = dU + d(PV) 5dh = dU + PdV + VdP 6But: dQ = dU + PdV dh = dQ + VdP 7From Eq. 3Q = h + KE + PE + W 8dQ = dh + dKE + dPE + dW 9dQ = dU + PdV + VdP + dKE + dPE + dW 10dQ = dQ + VdP + dKE + dPE + dW 0 = VdP + dKE + dPE + dW dW = -VdP - dKE - dPE 11By IntegrationW = - VdP - KE - PE 12
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If KE = 0 and PE = 0Q = h + W 13W = Q - h 14W = - VdP 15
PEKEhVdP-W
PEKEhQW
WPEKEhQ
SYSTEM OPEN an For .B
PdVW
dWdUdQ
WUQ
SYSTEM CLOSED a For .A
SUMMARY
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IDEAL OR PERFECT GAS
Prepared By: Engr Yuri G. Melliza
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1. Ideal Gas Equation of StatePV = mRTP = RT
2T2V
2P
1T1V
1P
CTPV
RTP
ρ
Where: P – absolute pressure in KPa V – volume in m3
m – mass in kg R – Gas Constant in KJ/kg-°K T – absolute temperature in°K
IDEAL OR PERFECT GAS
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2. Gas Constant
K-m
kgKJ
8.3143R
K-kgKJ
MR
R
Where:R- Gas Constant in KJ/kg-K
Km
kgKJ
constant gas universal R
M – Molecular weight kg/kgm
3. Boyle’s Law If the temperature of a certain quantity of gas is held constant the volume V is inver- sely proportional to the absolute pressure P.
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C2V
2P
1V
1P
CPVP1
CV
PV
α
4.Charle’s LawA. At Constant Pressure (P = C) If the pressure of a certain quantity ofgas is held constant, the volume V is directly proportional to the temperature T during a qua-sistatic change of state
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2
2
1
1
T
V
T
V
CTV
T;CV; T α V
B. At Constant Volume (V = C)If the volume of a certain quantity of gas isheld constant, the pressure P varies directlyas the absolute temperature T.
2
2
1
1
T
P
T
P
CTP
; TCPT α P
;
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5. Avogadro’s LawAll gases at the same temperature and
pressure have the same number of molecules per unit of volume, and it follows that thespecific weight is directly proportional toits molecular weight M.
M
6.Specific HeatSpecific Heat or Heat Capacity is the amountof heat required to raise the temperature of a 1 kg mass 1C or 1KA. SPECIFIC HEAT AT CONSTANT PRESSURE (Cp)
From: dh = dU + PdV + VdPbut dU + VdP = dQ ; therefore
dh = dQ + VdP 1
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but at P = C ; dP = O; thereforedh = dQ 2
and by integrationQ = h 3
considering m, h = m(h2 - h1) 4Q = h = m (h2 - h1) 5
From the definition of specific heat, C = dQ/TCp = dQ /dt 6
Cp = dh/dT, then dQ = CpdT 7
and by considering m,dQ = mCpdT 8
then by integration Q = m Cp T 9
but T = (T2 - T1)Q = m Cp (T2 - T1) 10
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B SPECIFIC HEAT AT CONSTANT VOLUME (Cv)At V = C, dV = O, and from dQ = dU + PdV dV = 0, therefore
dQ = dU 11 then by integration
Q = U 12then the specific heat at constant volume Cv is;
Cv = dQ/dT = dU/dT 13 dQ = CvdT 14
and by considering m, dQ = mCvdT 15
and by integration Q = mU 16Q = mCvT 17 Q = m(U2 - U1) 18 Q = m Cv(T2 - T1) 19
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From: h = U + P and P = RT h = U + RT 20
and by differentiation, dh = dU + Rdt 21 but dh =CpdT and dU = CvdT,
therefore CpdT = CvdT + RdT 22and by dividing both sides of the
equation by dT, Cp = Cv + R 23
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7. Ratio Of Specific Heatsk = Cp/Cv 24k = dh/du 25k = h/U 26
From eq. 32,Cp = kCv 27
substituting eq. 27 to eq. 24Cv = R/k-1 28
From eq. 24,
Cv = Cp/k 29substituting eq. 29 to eq. 24
Cp = Rk/k-1 30
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8. Entropy Change (S)Entropy is that property of a substance that determines the amount of randomness and disorder of a substance. If during a process, an amount of heat is taken and is by divided by the absolute temperature at which it is taken, the result iscalled the ENTROPY CHANGE.
dS = dQ/T 31and by integration
S = ∫dQ/T 32and from eq. 39
dQ = TdS 33
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2
2
1
1
2
2
1
1
2
2
1
1
2211
2
22
1
11
MM
LAW SAVOGADRO' .4
CTV
TV
C P At b.
CTP
TP
C V At a.
LAW CHARLES .3
CVPVP
C) T ( LAW BOYLES 2.
mRTPV
CTVP
TVP
State of Equation .1
SUMMARY
T
dQS
CHANGE ENTROPY .8CvCp
k
HEAT SPECIFIC OF RATIO .7
RCvCp 1-k
RCv ;
1-kRk
Cp
HEAT PECIFICS 6.
kgkg
R
8.3143M
K-kgKJ
M
8.3143R
CONSTANT GAS .5
mol
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GAS MIXTURE
Total Mass of a mixture
inn
mm
x ii
imm Mass Fraction
Total Moles of a mixture
nn
y ii
Mole Fraction
Where:m – total mass of a mixturemi – mass of a componentn – total moles of a mixtureni – moles of a componentxi – mass fraction of a componentyi - mole fraction of a component
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Equation of StateMass Basis
A. For the mixture
iiiii TRmVP
mRTPV
TRnPV
B. For the components
iiiii TRnVP
Mole Basis
A. For the mixture
B. For the components
Where:R – Gas constant of a mixture
in KJ/kg-K - universal gas constant in
KJ/kgm- KR
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AMAGAT’S LAW The total volume of a mixture V is equal to the volume occupied by each component at the mixture pressure P and temperature T.
1n1
V1
2n2
V2
3n3
V3
P,T
P = P1 = P2 = P3
T = T1 = T2 = T3
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For the components:
TR
PVn ;
TR
PVn ;
TR
PVn 3
32
21
1
The mole fraction:
V
Vyi
TR
PVTR
PV
y
n
ny
i
i
i
ii
321
321
321
321
VVVV
P
TR
TR
PV
TR
PV
TR
PV
TR
PV
TR
PV
TR
PV
TR
PV
TR
PV
nnnn
The total moles n:
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DALTON’S LAW The total pressure of a mixture P is equal to the sum of the partial pressure that each gas would exert at mixture volume V and temperature T.
1n1
P1
2n2
P2
3n3
P3
MIXTURE
nP
T1 = T2 = T3 = TV1 = V2 = V3 = V
For the mixture
For the components
TR
VPn
TR
VPn
TR
VPn
33
22
11
TR
PVn
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321
321
321
321
PPPP
V
TR
TR
VP
TR
VP
TR
VP
TR
PV
TR
VP
TR
VP
TR
VP
TR
PV
nnnn
The total moles n: The mole fraction:
P
Pyi
TR
PVTR
VP
y
n
ny
i
i
i
ii
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Molecular Weight of a mixture
R
RM
MyM ii
M
RR
RxR ii
Gas Constant of a mixture
Specific Heat of a mixture
RCC
CxC
CxC
vp
viiv
piip
Ratio of Specific Heat
u
h
C
Ck
v
p
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Gravimetric and Volumetric AnalysisGravimetric analysis gives the mass fractions of the components
in the mixture. Volumetric analysis gives the volumetric or molal fractionsof the components in the mixture.
i
i
i
i
i
ii
iii
Mx
Mx
y
My
Myx
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PROPERTIES OF PURE SUBSTANCE
a - sub-cooled liquidb - saturated liquidc - saturated mixtured - saturated vapore - superheated vapor
Considering that the system is heated at constant pressure where P = 101.325 KPa, the 100C is the saturation temperature corresponding to 101.325 KPa, and 101.325 KPa is the saturation pressure correspon-ding 100C.
P P P P P
Q
30°C100°C
100°C 100°CT100°C
(a) (b) (c) (d) (e)
Q Q Q Q
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Saturation Temperature (tsat) - is the highest temperature at a given pressure in which vaporization takes place.Saturation Pressure (Psat) - is the pressure corresponding to the temperature.Sub-cooled Liquid - is one whose temperature is less than the saturation temperature corresponding to the pressure.Compressed Liquid - is one whose pressure is greater than the saturation pressure corresponding to the temperature. Saturated Liquid - a liquid at the saturation temperatureSaturated Vapor - a vapor at the saturation temperatureSaturated Mixture - a mixture of liquid and vapor at the saturation temperature.Superheated Vapor - a vapor whose temperature is greater than the saturation temperature.
a
b c de
T
F
Saturated Vapor
Saturated Vapor
30°C
100°C
t 100°C
Saturated Mixture
P = C
Critical Point
T- Diagram
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a
b c de
T
S
F
Saturated Vapor
Saturated Vapor
30°C
100°C
t 100°C
Saturated Mixture
P = C
Critical Point
T-S Diagram
F(critical point)- at the critical point the temperature and pressure is unique.For Steam: At Critical Point, P = 22.09 MPa; t = 374.136C
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a
b c de
T
S
F
Saturated Vapor
Saturated Vapor
ta
tsat
te
Saturated Mixture
P = C
Critical Point
T-S Diagram
tsat - saturation temperature corresponding the pressure Pta - sub-cooled temperature which is less than tsatte - superheated vapor temperature that is greater than tsat
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h-S (Enthalpy-Entropy Diagram)
h
S
t = C (constant temperature curve)
P = C (constant pressure curve)F
I
II
III
I - subcooled or compressed liquid regionII - saturated mixture regionIII - superheated vapor region
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Quality (x):
Lv
v
Lv
v
mmm
m
m
mm
mx
Where:mv – mass of vapormL – mass of liquidm – total massx- quality
The properties at saturated liquid, saturated vapor, superheatedvapor and sub-cooled or compressed liquid can be determined from tables. But for the properties at saturated mixture (liquid and vapor) they can be determined by the equation
rc = rf + x(rfg) rfg = rg – rf
Where: r stands for any property (, U, h and S)rg – property at saturated vapor (from table)rf – property at saturated liquid
Note: The properties at siub-cooled or compressed liquid is approximately equal to the properties at saturated liquidcorresponding the sub-cooled temperature.
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Throttling Calorimeter
Main Steam Line
P1 – steam line pressure
To main steam line
P2 -Calorimeter pressure
h1 = h2
h1 = hf1 + x1(hfg1)Where:
1 – main steam line2 - calorimeter
thermometer
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P1
P2
1
2
T
S
h = C
T-S Diagram Throttling Process
P1 – steam line pressureP2 – calorimeter pressure
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1. Isobaric Process ( P = C): An Isobaric Process is an internally reversible constant pressure process. A. Closed System:(Nonflow)
P
V
21P
dV
Q = U + W 1 any substanceW = PdV 2 any substanceU = m(U2 - U1) 3 any substanceW = P(V2 - V1) 4 any substanceQ = h = m(h2-h1) 5 any substance
T
S
1
2
dS
TP = C
PROCESSES OF FLUIDS
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For Ideal Gas:PV = mRTW =mR(T2-T1) 5U = mCv(T2-T1) 6 Q = h = mCP (T2-T1) 7Entropy ChangeS = dQ/T 8 any substancedQ = dhFor Ideal Gasdh = mCPdTS = dQ/TS = mCP dT/TS = mCP ln(T2/T1) 9B. Open System:Q = h + KE + PE + W 10 any substanceW = - VdP - KE - PE 11 any substance- VdP = 0
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Q = h 12W = - KE - PE 13If KE = 0 and PE = 0W = 0 14Q = mCP(T2-T1) 15 Ideal Gas
2. Isometric Process (V = C): An Isometric process is internally reversible constant volume process.
A. Closed System: (Nonflow)
P
V1
2T
S
T
dS
1
2V = C
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Q = U + W 1 any substanceW = PdV at V = C; dV = 0W = 0 Q = U = m(U2 - U1) 2 any substanceh = m(h2-h1) 3 any substance
For Ideal Gas:Q = U = mCv(T2-T1) 4 h = mCP(T2-T1) 5Entropy Change:S = dQ/T 6 any substancedQ = dUdU = mCvdT for ideal gasS = dU/T = mCvdT/TS = mCv ln(T2/T1) 6
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B. Open System:Q = h + KE + PE + W 7 any substanceW = - VdP - KE - PE 8 any substance-VdP = -V(P2-P1) 9 any substanceQ = U = m(U2 - U1) 10 any substanceh = m(h2-h1) 11 any substanceFor Ideal Gas:-VdP = -V(P2-P1) = mR(T1-T2)Q = U = mCv(T2-T1) 12 h = mCP(T2-T1) 13If KE = 0 and PE = 0Q = h + W 14 any substanceW = - VdP 15W = -VdP = -V(P2-P1) 16 any substance W = mR(T1-T2) 16 ideal gash = mCP(T2-T1) 17 ideal gas
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3. Isothermal Process(T = C): An Isothermal process is a reversible constant temperature process. A. Closed System (Nonflow)
dS
T
S
T1 2
P
V
1
2P
dV
PV = C orT = C
Q = U + W 1 any substanceW = PdV 2 any substanceU = m(U2 - U1) 3 any substanceFor Ideal Gas:dU = mCv dT; at T = C ; dT = 0Q = W 4
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W = PdV ; at PV = C ; P1V1 = P2V2 = C; P = C/VSubstituting P = C/V to W = PdV W = P1V1 ln(V2/V1) 5Where (V2/V1) = P1/P2
W = P1V1 ln(P1/P2) 6P1V1 = mRT1 Entropy Change:dS = dQ/T 7S = dQ/TdQ = TdS ;at T = CQ = T(S2-S1)(S2-S1) = S = Q/T 8S = Q/T = W/T 9 For Ideal Gas
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B. Open System (Steady Flow)Q = h + KE + PE + W 10 any substanceW = - VdP - KE - PE 11 any substance-VdP = -V(P2-P1) 12 any substanceh = m(h2-h1) 13 any substanceFor Ideal Gas:-VdP = -P1V1ln(P2/P1) 14 -VdP = P1V1ln(P1/P2) 15 P1/P2 = V2/V1 16dh = CPdT; at T = C; dT = 0h = 0 16 If KE = 0 and PE = 0Q = h + W 17 any substanceW = - VdP = P1V1ln(P1/P2) 18For Ideal Gash = 0 19Q = W = - VdP = P1V1ln(P1/P2) 20
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4. Isentropic Process (S = C): An Isentropic Process is an internally“Reversible Adiabatic” process in which the entropy remains constantwhere S = C and PVk = C for an ideal or perfect gas.
For Ideal Gas
1
2
1
1
1
2
1
2
2
22
1
kk
k
k22
k11
11
k
V
V
P
P
T
T
VPVP and T
VP
T
VP
C PV and CTPV Using
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A. Closed System (Nonflow)
T
S
1
2
P
V
1
2
dV
P
S = C orPVk = C
Q = U + W 1 any substanceW = PdV 2 any substanceU = m(U2 - U1) 3 any substanceQ = 0 4W = - U = U = -m(U2 - U1) 5
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For Ideal Gas U = mCV(T2-T1) 6 From PVk = C, P =C/Vk, and substituting P =C/Vk to W = ∫PdV, then by integration,
11
11
1
1
1
211
1
1
21
kk
VP
kk
12
1122
P
P
kPdV
P
P
k
mRT
k-1
T-TmRPdV
k
VP-VPPdV W 7
8
9
Q = 0
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Entropy ChangeS = 0S1 = S2
B. Open System (Steady Flow)Q = h + KE + PE + W 10 any substanceW = - VdP - KE - PE 11 any substanceh = m(h2-h1) 12 any substanceQ = 0W = -h - KE - PE 13From PVk = C ,V =[C/P]1/k, substituting V to-∫VdP, then by integration,
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11
11
1
1
1
211
1
1
21
kk
kk
12
1122
P
P
k
VkPVdP
P
P
k
kmRT
k-1
T-TkmRVdP
k
VP-VPkVdP
PdV kVdP
14
15
16
If KE = 0 and PE = 00 = h + W 17 any substanceW = - VdP = - h 18 any substanceh = m(h2-h1) 19 any substanceQ = 0
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12P
kk
kk
12
1122
T-TmChW
P
P
k
VkPW
P
P
k
kmRT
k-1
T-TkmRW
k
VP-VPkPdV kVdPW
11
11
1
1
1
211
1
1
21
20
22
21
23
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P
V
dP
V
Area = -VdP
S = C
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1
2
1
1
1
2
1
2
2
22
1
nn
n
n22
n11
11
n
V
V
P
P
T
T
VPVP and T
VP
T
VP
C PV and CTPV Using
5. Polytropic Process ( PVn = C): A Polytropic Process is an internally reversible process of an Ideal or Perfect Gas in which PVn = C, where n stands for any constant.
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A. Closed System: (Nonflow)
Q = U + W 1 W = PdV 2 U = m(U2 - U1) 3 Q = mCn(T2-T1) 4U = m(U2 - U1) 5
P
V
1
2
dV
P
PVn = C
T
S
2
1
dS
T
PVn = C
K-kgKJ
or C-kg
KJ heat specific polytropic C
n1nk
CC
n
vn
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From PVn = C, P =C/Vn, and substituting P =C/Vn to W = ∫PdV, then by integration,
11
11
1
1
1
211
1
1
21
nn
VP
nn
12
1122
P
P
nPdVW
P
P
n
mRT
n-1
T-TmRPdVW
n
VP-VPPdV W
Entropy ChangedS = dQ/TdQ = mCndTS = mCnln(T2/T1)
6
8
9
10
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B. Open System (Steady Flow)Q = h + KE + PE + W 11 W = - VdP - KE - PE 12 h = m(h2-h1) 13 Q = mCn(T2-T1) 14dQ = mCn dTW = Q - h - KE - PE 15From PVn = C ,V =[C/P]1/n, substituting V to-∫VdP, then by integration,
n
VP-VPnVdP
PdV nVdP
1122
1 16
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11
11
1
1
211
1
1
21
nn
nn
12
P
P
n
VnPVdP
P
P
n
nmRT
n-1
T-TnmRVdP
If KE = 0 and PE = 0Q = h + W 19 any substanceW = - VdP = Q - h 20 any substanceh = m(h2-h1) 21 any substanceh = mCp(T2-T1)Q = mCn(T2-T1) 22
17
18
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11
11
1
1
211
1
1
21
nn
nn
12
P
P
n
VnPW
P
P
n
nmRT
n-1
T-TnmRW
24
23
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6. Isoenthalpic or Throttling Process: It is a steady - state, steady flow process in which Q = 0; PE = 0; KE = 0; W = 0 and the enthalpy remains constant.
h1 = h2 or h = C
Throttling valve
Main steam line
thermometer
Pressure Gauge
Pressure Gauge
To main steam line
Throttling Calorimeter
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Irreversible or Paddle Work
m
W
Q
UWp
Q = U + W - Wp
where: Wp - irreversible or paddle work
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THERMODYNAMICS Part 2By. Engr. Yuri G. Melliza
2nd Law of Thermodynamics
Carnot Cycles
Steam Cycles
Fuels and Combustion
ICE Cycles
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2nd Law of Thermodynamics
• Second Law of Thermodynamics• Kelvin – Planck Statement• Carnot engine• Carnot Refrigerator• Sample Problems
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Second Law of Thermodynamics:Whenever energy is transferred, the level of energy cannot be conserved and some energy must be permanently reduced to a lower level. When this is combined with the first law of thermodynamics, the law of energy conservation, the statement becomes:
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Second Law of Thermodynamics:Whenever energy is transferred, energy must be conserved, but the level of energy cannot be conserved and some energy must be permanently reduced to a lower level.
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Kelvin-Planck statement of the Second Law:No cyclic process is possible whose sole result is the flow of heat from a single heat reservoir and the performance of an equivalent amount of work.For a system undergoing a cycle: The net heat is equal to the net work. QW dWdQ Where:
W - net workQ - net heat
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CARNOT CYCLENicolas Leonard Sadi Carnot 1796-1832 1.Carnot Engine Processes:
1 to 2 - Heat Addition (T = C)2 to 3 - Expansion (S = C)3 to 4 - Heat Rejection (T = C)4 to 1 - Compression (S = C)
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P
V
2
1
3
4
T = C
S = CS = C
T = C
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T
S
21
34
T H
T L
Q A
Q R
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Heat Added (T = C)
QA = TH(S) 1
Heat Rejected (T = C)QR = TL(S) 2S = S2 - S1 = S3 – S4 3
Net WorkW = Q = QA - QR 4W = (TH - TL)(S) 5
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% xQ
Qe
% xQ
QQe
% x QW
e
A
R
A
RA
A
1001
100
100
6
7
8
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Substituting eq.1 and eq. 5 to eq 6
% xT
Te
% x T
TTe
H
L
H
LH
1001
100
9
10
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TH
TL
W
QA
QR
E
Carnot Engine
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2. Carnot Refrigerator: Reversed Carnot CycleProcesses: 1 to 2 - Compression (S =C) 2 to 3 - Heat Rejection (T = C) 3 to 4 - Expansion (S = C) 4 to 1 - Heat Addition (T = C)
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Q
R
Q
A
1
2
4
3
S
T
TH
TL
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Heat Added (T = C)QA = TL(S) 1
Heat Rejected (T = C)QR = TH(S) 2S = S1 - S4 = S2 - S3 3
Net WorkW = Q 4
W = QR - QA 5W = (TH - TL)(S) 6
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Coefficient of Performance(COP)
LH
L
A
TT
TCOP
W
QCOP
7
8
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1H
L
T
TCOP 9
Tons of Refrigeration211 KJ/min = 1 TR3. Carnot Heat Pump:A heat pump uses the same components as therefrigerator but its purpose isto reject heat at high energy level.
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Performance Factor (PF)
AR
R
R
QPF
W
QPF
10
11
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1
1
1
COPPF
T
TPF
Q
QPF
TT
TPF
L
H
A
R
LH
H 12
13
14
15
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TH
TL
W
QA
QR
R
Carnot Refrigerator
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A Carnot engine operating between 775 K and 305K produces 54 KJ of work. Determine the change of entropy during heat addition.TH = 775 K ; TL = 305 KW = 54 KJ
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TH
TL
W
QA
QR
E
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K
KJ 0.015
775
89.04
T
QS-S
)S-(STQ
KJ 89.040.606
54
e
WQ
Q
We
0.606775
305775
T
TTe
H
A12
12HA
A
A
H
LH
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A Carnot heat engine rejects 230 KJ of heat at 25C. The net cycle work is 375 KJ. Determine the cycle thermal efficiency and the cycle high temperature .Given:QR = 230 KJTL = 25 + 273 = 298KW = 375 KJ
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TL = 298K
TH
WE
QR = 230 KJ
QA
K87.783772.0
605
)S-(S
QT
KKJ/ -0.772)S-(S
KKJ/ 772.0)SS(
)SS(298230
)SS(SS
)SS(TQ
)SS(TQ
62.0605
375
QA
We
KJ 605QA
)230375(QWQ
QQW
12
AH
12
34
34
1234
34LR
12HA
RA
RA
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A Carnot engine operates between temperature reservoirs of 817C and 25C and rejects 25 KW to the low temperature reservoir. The Carnot engine drives the compressor of an ideal vapor compres-sion refrigerator, which operates within pressure limits of 190 KPa and 1200 Kpa. The refrigerant is ammonia. Determine the COP and the refrigerant flow rate.(4; 14.64 kg/min)TH = 817 + 273 = 1090 KTL = 25 + 273 = 298 KQR = 25 KW
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Internal Combustion Engine Cycles
1. Air Standard Otto Cycle (Spark Ignition Engine Cycle)Processes1 to 2 - Isentropic Compression (S = C)2 to 3 - Constant Volume Heat Addition ( V = C)3 to 4 - Isentropic Expansion (S =C)4 to 1 - Constant Volume Heat Rejection (V = C)
P
PmV
VD
CVD
W1
42
3
S = C
S = C
3QAT
S
1
24
V = C
V = C
QR
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Compression Ratio
3
4
2
1
V
V
V
Vr
where: r - compression ratioV1 = V4 and V2 = V3
1
Heat Added (V = C) QA = mCV(T3 - T2) 2
Heat RejectedQR = mCV(T4 - T1) 3
Net Cycle WorkW = QA - QR
W = mCV[(T3 - T2) - (T4 - T1)] 4
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Thermal Efficiency
100%r
11e
100%TT
TT1e
100%x Q
Q1e
100%Q
QQe
100%xQ
We
1-k
23
14
A
R
A
RA
A
x
x
x
5
6
7
8
9
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Mean Effective Pressure
KPa V
WP
Dm
where:W - net work, KJ, KJ/kg, KWVD - Displacement Volume, m3, m3/kg, m3/secVD = V1 - V2 m3
VD = 1 - 2 m3/kg
10
Percent Clearance
100% x V
VC
D
2
V2 = CVD
V1 = Vd + CVD
C
C1
V
Vr
2
1 11 12
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2. Diesel Cycle: (Compression Ignition Engine Cycle)
Processes1 to 2 - Isentropic Compression (S = C)2 to 3 - Constant Pressure Heat Addition (P = C)3 to 4 - Isentropic Expansion (S = C)4 to 1 - Constant Volume Heat Rejection (V = C)
P
V
T
S
2 3
4
1
S = C
S = C
VDCVD
1
2
3
4V = C
P = C
QR
QA
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Heat Added (P = C) QA = mCP(T3 - T2) 3 QA = mkCV(T3 - T2) 4
Heat Rejected (V = C)QR = mCV(T4 - T1) 5
Net Cycle WorkW = QA - QR
W = mCV[k(T3 - T2) - (T4 - T1)] 6
Compression Ratio
3
4
2
1
V
V
V
Vr 1
Cut - Off Ratio
2
3c V
Vr 2
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Thermal Efficiency
100%1)k(r
1)(r
r
11e
100%TT
TT1e
100%x Q
Q1e
100%Q
QQe
100%xQ
We
c
kc
1-k
23
14
A
R
A
RA
A
x
x k
x
7
8
9
10
11
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where:W - net work, KJ, KJ/kg, KWVD - Displacement Volume, m3, m3/kg, m3/secVD = V1 - V2 m3
VD = 1 - 2 m3/kg
KPa V
WP
Dm
Mean Effective Pressure
12
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3. Air Standard Dual CycleProcesses:1 to 2 - Compression (S = C)2 to 3 -Heat Addition (V = C)3 to 4 - Heat Addition (P = C)4 to 5 -Expansion (S = C)5 to 1 _ Heat Rejection (V = C)
11
2
2
33
44
5
5S = C
S = C
P = C
V = C
V = C
QA1
QA2
QR
P
V
T
S
VDCVD
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Comprssion Ratio
4
5
3
5
2
1
V
V
V
V
V
Vr
V5 = V1 ; V2 = V3
1
Pressure Ratio
2P
4P
2
3p P
Pr 3
2
4
3
4c V
Vr
V
V 2
Cut-Off Ratio
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Heat Added (V = C) QA1 = mCV(T3 - T2) 4Heat Added (P =C)QA2 = mCP(T4 - T3) 5QA2 = mkCV(T4 - T3) 6
Heat Rejected (V = C)QR = mCV(T5 - T1) 7
Net Cycle WorkW = QA - QR 8QA = QA1 + QA2 9QA = mCV[(T3 - T2) +k (T4 - T3)] 10W = mCV[(T3 - T2) + k(T4 - T3) - (T5 - T1) ]
100%x Q
Q1e
100%Q
QQe; 100%x
Q
We
A
R
A
RA
A
x
12
13
Thermal Efficiency
11
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100%1-(rkr1)-(r
1)-r(r
r
11e
100% )]T - (Tk )T - [(T
)T - (T1e
cpp
kcp
1-k
3423
15
x
x
Mean Effective PressurePm = W/VD KPa
where: VD = V1 - V2 m3 ; W in KJ
V1 = V5 ; V2 = V3
VD = 1 - 2 m3/kg ; W in KJ/kg
1 = 5 ; 2 = 3
For Cold Air Standard: K = 1.4For Hot Air Standard: K = 1.3
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Vapor Power Cycle
RANKINE CYCLEProcesses:
1 to 2 - Expansion (S = C)2 to 3 - Heat Rejection (P = C)3 to 4 - Compression or Pumping (S = C)4 to 1 - Heat Addition (P = C)
Boiler or SteamGenerator
Turbine
Condenser
Pump
WP
QA
QR
Wt
1
2
3
4
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Major Components of a Rankine Cycle1. Steam Generator or Boiler: The working substance absorbs heat from products of combustion or other sources of heat at constant pressure which in turn changes the state of the working substance (water or steam) from sub-cooled liquid and finally to superheated vapor whence at this point it enters the turbine. 2. Steam Turbine: A steady state, steady flow device where steam expands isentropically to a lower pressure converting some forms of energy (h, KE, PE) to mechanical work that finally be converted into electrical energy if the turbine is used to drive an electric gene- rator.3. Condenser: Steam exiting from the turbine enters this device to re- ject heat to the cooling medium and changes its state to that of the saturated liquid at the condenser pressure which occurred at a cons- tant pressure process.
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4. Pump: It is also a steady state, steady flow machine where the condensate leaving the condenser at lower pressure be pumped back to the boiler in an isentropic process in order to raise the pressure of the condensate to that of the boiler pressure.
h
S S
T
3
42
1
3
4
1
2
P1
P2
P1
P2
4’2’
2’
4’
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Turbine Worka) Ideal Cycle
Wt = (h1 - h2) KJ/kgWt = ms(h1 - h2) KW
b) Actual CycleWt’ = (h1 - h2’) KJ/kgWt’ = ms(h1 - h2’) KW
where: ms - steam flow rate in kg/secTurbine Efficiency
100%x hhhh
η
100%x WW
tη
21
2'1t
t
t'
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Pump Worka) Ideal Cycle
WP = (h4 - h3) KJ/kgWP = ms(h4 - h3) KW
b) Actual CycleWP’ = (h4’ - h3) KJ/kgWP’ = ms(h4’ - h3) KW
Pump Efficiency
100%xhh
hhη
100%xW
Wη
34'
34p
p'
pp
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Heat Rejecteda) Ideal Cycle
QR = (h2 - h3) KJ/kgQR = ms(h2 - h3) KW QR = ms(h2 - h3) KW = mwCpw(two - twi) KW
b) Actual CycleQR = (h2’ - h3) KJ/kgQR = ms(h2’ - h3) KW = mwCpw(two - twi) KW
Where: mw - cooling water flow rate in kg/sec twi - inlet temperature of cooling water inC two - outlet temperature of cooling water inC Cpw - specific heat of water in KJ/kg- C or KJ/kg-K Cpw = 4.187 KJ/kg- C or KJ/kg- K
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Heat Added:a) Ideal Cycle
QA = (h1 - h4) KJ/kgQA = ms (h1 - h4) KW
b) Actual CycleQA = (h1 - h4’) KJ/kgQA = ms (h1 - h4’) KW
Steam Generator or boiler Efficiency
100%x(HV)m
)h(hmη
100%xQ
Qη
f
41sB
S
AB
Where: QA - heat absorbed by boiler in KWQS - heat supplied in KWmf - fuel consumption in kg/secHV - heating value of fuel in KJ/kg
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Steam Rate
KW-sec
kg
ProducedKW
rate Flow SteamSR
Heat Rate
KW-sec
KJ
ProducedKW
SuppliedHeat HR
Reheat Cycle A steam power plant operating on a reheat cycle improves the thermalefficiency of a simple Rankine cycle plant. After partial expansion of the steam in the turbine, the steam flows back to a section in the boiler which is the re-heater and it will be reheated almost the same to its initial temperature and expands finally in the turbine to the con-denser pressure.
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Reheater
QA
WP
QR
Wt
1 kg
12 3
4
56
Regenerative Cycle In a regenerative cycle, after partial expansion of the steam in theturbine, some part of it is extracted for feed-water heating in an open orclose type feed-water heater. The bled steam heats the condensate from the condenser or drains from the previous heater causing a decrease in heat absorbed by steam in the boiler which result to an increase in thermal efficiency of the cycle.
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QA
WP1
QR
Wt
1 kg
1
2
3
456
7
WP2
m
Reheat-Regenerative Cycle For a reheat - regenerative cycle power plant, part of the steam is re-heated in the re-heater and some portion is bled for feed-water heating to an open or closed type heaters after its partial expansion in the turbine. It will result to a further increase in thermal efficiency of theplant.
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QA
WP1
QR
Wt
1 kg
1
2
4
5678
WP2
m
23
1-m
1-m
For a 1 kg basis of circulating steam, m is the fraction of steam extracted for feed-water heating as shown on the schematic diagram above, where the reheat and bled steam pressure are the same.
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FUELS and
COMBUSTION
By. Engr. Yuri G. Melliza
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FUELS AND COMBUSTION Fuels and Combustion Types of Fuels Complete/Incomplete Combustion Oxidation of Carbon Oxidation of Hydrogen Oxidation of Sulfur Air composition Combustion with Air Theoretical Air Hydrocarbon fuels Combustion of Hydrocarbon Fuel
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Fuels and Combustion
Fuel: Substance composed of chemical elements which in rapid chemical union with oxygen produced combustion.
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Combustion: Is that rapid chemical union with oxygen of an element, whose exothermic heat of reaction is sufficiently great and whose rate of reaction is suffi-ciently fast whereby useful quantities of heat are liberated at elevated temperature.
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TYPES OF FUELS Solid Fuels
ex: Wood, coal, charcoal Liquid Fuels
ex: gasoline, diesel, kerosene Gaseous Fuels
ex: LPG, Natural Gas, Methane Nuclear Fuels
ex: Uranium
Combustible Elements1. Carbon (C) 3. Sulfur (S)2. Hydrogen (H2)
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Complete Combustion: Occurs when all the combustible elements has been fully oxidized.
Ex:C + O2 CO2
Incomplete Combustion: Occurs when some of the combustible elements has not been fully oxidized.
Ex:C + O2 CO
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Common Combustion GasesGAS MOLECULAR
Weight (M)
C 12
H 1
H2 2
O 16
O2 32
N 14
N2 28
S 32
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THE COMBUSTION CHEMISTRY Oxidation of Carbon
11 83
44 3612
32)1(12 1(16)1(12)
Basis Mass
1 11
Basis Mole
CO OC 2
2
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Oxidation of Hydrogen
9 81
18 162
2)1(16 (32)1(2)
Basis Mass
1 1
Basis Mole
OH OH
2
1
21
2
22 21
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Oxidation of Sulfur
2 11
64 3232
32)1(32 (32)1(32)
Basis Mass
1 11
Basis Mole
OS OS
1
22
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Composition of AIRa. Percentages by Volume (by
mole)O2 = 21%N2 = 79%
b. Percentages by MassO2 = 23%N2 = 77%
76321
79.
2
2
O of MoleN of Moles
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Combustion with AirA. Combustion of Carbon with air
C + O2 + 3.76N2 CO2 + 3.76N2
Mole Basis:1 + 1 + 3.76 1+ 3.76
Mass Basis:1(12) + 1(32) + 3.76(28) 1(44) + 3.76(28)12 + 32 + 3.76(28) 44 + 3.76(28) 3 + 8 + 3.76(7) 11+ 3.76(7)
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kg of air per kg of Carbon:
C of kgair of kg
11.44=3
3.76(7)+8=
C of kgair of kg
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B. Combustion of Hydrogen with air H2 + ½ O2 + ½ (3.76)N2 H2O + ½(3.76)N2 Mole Basis:
1 + ½ + ½(3.76) 1 + ½(3.76)Mass Basis:
1(2) + ½ (32) + ½(3.76)(28) 1(18) + ½ (3.76)(28)
2 + 16 + 3.76(14) 18 + 3.76(14) 1 + 8 + 3.76(7) 9 + 3.76(7)
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kg of air per kg of Hydrogen:
22 H of kgair of kg
34.32=1
3.76(7)+8=
H of kgair of kg
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C. Combustion of Sulfur with airS + O2 + 3.76N2 SO2 + 3.76N2
Mole Basis:1 + 1 + 3.76 1 + 3.76
Mass Basis:1(32) + 1(32) + 3.76(28) 1(64) +
3.76(28) 32 + 32 + 105.28 64 + 105.28
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kg of air per kg of Sulfur:
S of kgair of kg
4.29=32105.2832
=S of kgair of kg
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Theoretical AirIt is the minimum amount of air required to oxidize the reactants or the combustible elements found in the fuel. With theoretical air no O2 is found in products.
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Excess AirIt is an amount of air in excess of the Theoretical requirements in order to influence complete combustion. With excess air O2 is present in the products.
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HYDROCARBON FUELSFuels containing the element s Carbon and Hydrogen. Chemical Formula: CnHm
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Family Formula Structure Saturated
Paraffin CnH2n+2 Chain Yes
Olefin CnH2n Chain No
Diolefin CnH2n-2 Chain No
Naphthene CnH2n Ring Yes
Aromatic
Benzene CnH2n-6 Ring No
Naphthalene CnH2n-12 Ring No
Alcohols Note: Alcohols are not pure hydrocarbon, because one of its hydrogen atom is replace by an OH radical. Sometimes it is used as fuel in an ICE.
Methanol CH3OH
Ethanol C2H5OH
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Saturated Hydrocarbon: All the carbon atoms are joined by a single bond.Unsaturated Hydrocarbon: It has two or more adjacent Carbon atoms joined by a double or triple bond.Isomers: Two hydrocarbons with the same number of carbon and hydrogen atoms but atdifferent structures.
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H H H H H C C C CH H H H H
Chain structure Saturated
H H HC C=C C H H H H H
Chain Structure Unsaturated
Ring structure Saturated H H H C H C C H C H H H
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Theoretical Air: It is the minimum or theoretical amount of air required to oxidized the reactants. With theoretical air no O2 is found in the products. Excess Air: It is an amount of air in excess of the theo-retical air required to influence complete combustion. With excess air O2 is found in the products.
Combustion of Hydrocarbon Fuel(CnHm)
A. Combustion with 100% theoretical air CnHm + aO2 + a(3.76)N2 bCO2 + cH2O + a(3.76)N2
fuel
air
t kg kg
m12n
)a(3.76)(28a(32)FA
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Combustion of Hydrocarbon FuelFormula: (CnHm)
A. Combustion with 100% theoretical air CnHm + aO2 + a(3.76)N2 bCO2 + cH2O + a(3.76)N2
fuel
air
t kg kg
m12n
)a(3.76)(28a(32)FA
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fuel
air
a kg kg
m12n
)a(3.76)(28a(32)e)(1
FA
B. Combustion with excess air e CnHm +(1+e) aO2 + (1+e)a(3.76)N2 bCO2 +
cH2O + dO2 + (1+e)a(3.76)N2
Actual Air – Fuel Ratio
fuel
air
ta kg kg
FA
e)(1FA
Where: e – excess air in decimalNote: Sometimes excess air is expressible in terms of theoretical air. Example: 25% excess air = 125% theoretical air
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Orsat Analysis: Orsat analysis gives the volumetric or molal analysis of the PRODUCTS on a DRY BASIS, (no amount of H2O given).
Proximate Analysis: Proximate analysis gives the amount of Fixed Carbon, Volatiles, Ash and Moisture, in percent by mass. Volatiles are those compounds that evaporates at low temperature when the solid fuel is heated.
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ULTIMATE ANALYSIS: Ultimate analysis gives the amount of C, H, O, N, S in percentages by mass, and sometimes the amount of moisture and ash are given.
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SOLID FUELSComponents of Solid Fuels:
1. Carbon (C) 2. Hydrogen (H2)3. Oxygen (O2)4. Nitrogen (N2)5. Sulfur (S)6. Moisture (M)7. Ash (A)
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A.Combustion with 100% theoretical airaC + bH2 + cO2 + dN2 + eS + fH2O + gO2 + g(3.76)N2 hCO2 + iH2O + jSO2 + kN2
B.Combustion with excess air x: aC + bH2 + cO2 + dN2 + eS + fH2O +
(1+x)gO2 +(1+x)g(3.76)N2 hCO2 + iH2O + jSO2 + lO2 + mN2
WHERE: a, b, c, d, e, f, g, h, I, j, k, x are the number of moles of the elements.x – excess air in decimal
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fuel kgair kg
18f32e28d32c2b12a
3.76(28)g32gFA
t
Theoretical air-fuel ratio:
Actual air-fuel ratio:
fuel kgair kg
18f32e28d32c2b12a
3.76(28)g32gx)(1
a
F
A
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MASS FLOW RATE OF FLUE GAS (Products)
Air +Fuel Products
A. Without considering Ash loss
1
F
Amm Fg
B. Considering Ash loss
lossAsh 1
F
Amm Fg
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Heating Value
Heating Value - is the energy released by fuel when it is completely burned and the products of combustion are cooled to the original fuel temperature.Higher Heating Value (HHV) - is the heating value obtained when the water in the products is liquid.Lower Heating Value (LHV) - is the heating value obtained when the water in the products is vapor.
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For Solid Fuels with the presence of Fuel’s ULTIMATE ANALYSIS
kg
KJ S9304
8
OH212,144C820,33HHV 2
2
where: C, H2, O2, and S are in decimals from the ultimate analysis
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HHV = 31 405C + 141 647H KJ/kgHHV = 43 385 + 93(Be - 10) KJ/kg
For Liquid Fuels
where: Be - degrees Baume
For Coal and Oils with the absence of Ultimate Analysis
fuel of kg
air of Kg
3041
HHV
F
A
t
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For Gasoline
kgKJ )API(93639,38LHV
kgKJ )API(93160,41HHV
kgKJ )API(93035,39LHV
kgKJ )API(93943,41HHV
For Kerosene
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For Fuel Oils
Institute Petroleum AmericanAPI
kgKJ )API(6.139105,38LHV
kgKJ )API(6.139130,41HHV
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For Fuel Oils (From Bureau of Standard Formula)
).t(.St@S 561500070API131.5
141.5S
HHV = 51,716 – 8,793.8 (S)2 KJ/kgLHV = HHV - QL KJ/kg
QL = 2442.7(9H2) KJ/kg
H2 = 0.26 - 0.15(S) kg of H2/ kg of
fuel
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WhereS - specific gravity of fuel oil at 15.56 CH2 - hydrogen content of fuel oilQL - heat required to evaporate and superheat the water vapor formed bythe combustion of hydrogen in the fuelS @ t - specific gravity of fuel oil at any temperature tOxygen Bomb Calorimeter - instrument used in mea-suring heating value of solid and liquid fuels.Gas Calorimeter - instrument used for measuring heating value of gaseous fuels.
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Properties of Fuels and Lubricantsa) Viscosity - a measure of the resistance to flow that a lubricant offers when it is subjected to shear stress.b) Absolute Viscosity - viscosity which is determined by direct measurement of shear resistance.c) Kinematics Viscosity - the ratio of the absolute viscosity to the densityd) Viscosity Index - the rate at which viscosity changes with temperature.e) Flash Point - the temperature at which the vapor above a volatile liquid forms a combustible mixture with air.f) Fire Point - The temperature at which oil gives off vapor that burns continuously when ignited.
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g) Pour Point - the temperature at which oil will no longer pour freely.h) Dropping Point - the temperature at which grease melts.i) Condradson Number(carbon residue) - the percentage amount by mass of the carbonaceous residue remaining after destructive distillation.j) Octane Number - a number that provides a measure of the ability of a fuel to resist knocking when it is burnt in a gasoline engine. It is the percentage by volume of iso-octane in a blend with normal heptane that
matches the knocking behavior of the fuel.
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k) Cetane Number - a number that provides a measure of the ignition characteristics of a diesel fuel when it is burnt in a standard diesel engine. It is the percentage of cetane in the standard fuel.
Prepared By: ENGR YURI G. MELLIZA, RME